3.3.73 \(\int \frac {e^{-3 \tanh ^{-1}(a x)}}{\sqrt {c-a c x}} \, dx\) [273]

Optimal. Leaf size=67 \[ -\frac {8 \sqrt {c-a c x}}{a c \sqrt {1-a^2 x^2}}+\frac {2 (c-a c x)^{3/2}}{a c^2 \sqrt {1-a^2 x^2}} \]

[Out]

2*(-a*c*x+c)^(3/2)/a/c^2/(-a^2*x^2+1)^(1/2)-8*(-a*c*x+c)^(1/2)/a/c/(-a^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6262, 671, 663} \begin {gather*} \frac {2 (c-a c x)^{3/2}}{a c^2 \sqrt {1-a^2 x^2}}-\frac {8 \sqrt {c-a c x}}{a c \sqrt {1-a^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcTanh[a*x])*Sqrt[c - a*c*x]),x]

[Out]

(-8*Sqrt[c - a*c*x])/(a*c*Sqrt[1 - a^2*x^2]) + (2*(c - a*c*x)^(3/2))/(a*c^2*Sqrt[1 - a^2*x^2])

Rule 663

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*((a + c*x^2)^(p
+ 1)/(c*(p + 1))), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p,
0]

Rule 671

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*((a + c*x^2)^(p
+ 1)/(c*(m + 2*p + 1))), x] + Dist[2*c*d*(Simplify[m + p]/(c*(m + 2*p + 1))), Int[(d + e*x)^(m - 1)*(a + c*x^
2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p]
, 0]

Rule 6262

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> Dist[c^n, Int[(c + d*x)^(p - n)*(1 -
a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[a*c + d, 0] && IntegerQ[(n - 1)/2] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {e^{-3 \tanh ^{-1}(a x)}}{\sqrt {c-a c x}} \, dx &=\frac {\int \frac {(c-a c x)^{5/2}}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{c^3}\\ &=\frac {2 (c-a c x)^{3/2}}{a c^2 \sqrt {1-a^2 x^2}}+\frac {4 \int \frac {(c-a c x)^{3/2}}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{c^2}\\ &=-\frac {8 \sqrt {c-a c x}}{a c \sqrt {1-a^2 x^2}}+\frac {2 (c-a c x)^{3/2}}{a c^2 \sqrt {1-a^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 40, normalized size = 0.60 \begin {gather*} -\frac {2 \sqrt {1-a x} (3+a x)}{a \sqrt {1+a x} \sqrt {c-a c x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(3*ArcTanh[a*x])*Sqrt[c - a*c*x]),x]

[Out]

(-2*Sqrt[1 - a*x]*(3 + a*x))/(a*Sqrt[1 + a*x]*Sqrt[c - a*c*x])

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Maple [A]
time = 1.29, size = 50, normalized size = 0.75

method	result	size
gosper	\(\frac {2 \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}} \left (a x +3\right )}{\sqrt {-c x a +c}\, \left (a x +1\right )^{2} a \left (a x -1\right )}\)	\(46\)
default	\(\frac {2 \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}\, \left (a x +3\right )}{\left (a x -1\right ) \left (a x +1\right ) c a}\)	\(50\)
risch	\(\frac {2 \left (a x +1\right ) \sqrt {-\frac {\left (-a^{2} x^{2}+1\right ) c}{a x -1}}\, \left (a x -1\right )}{a \sqrt {\left (a x +1\right ) c}\, \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}}+\frac {4 \sqrt {-\frac {\left (-a^{2} x^{2}+1\right ) c}{a x -1}}\, \left (a x -1\right )}{\sqrt {\left (a x +1\right ) c}\, a \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}}\)	\(133\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2*(-a^2*x^2+1)^(1/2)*(-c*(a*x-1))^(1/2)*(a*x+3)/(a*x-1)/(a*x+1)/c/a

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Maxima [A]
time = 0.27, size = 30, normalized size = 0.45 \begin {gather*} -\frac {2 \, {\left (a x + 3\right )} \sqrt {a x + 1}}{a^{2} \sqrt {c} x + a \sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(1/2),x, algorithm="maxima")

[Out]

-2*(a*x + 3)*sqrt(a*x + 1)/(a^2*sqrt(c)*x + a*sqrt(c))

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Fricas [A]
time = 0.32, size = 43, normalized size = 0.64 \begin {gather*} \frac {2 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} {\left (a x + 3\right )}}{a^{3} c x^{2} - a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(1/2),x, algorithm="fricas")

[Out]

2*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*(a*x + 3)/(a^3*c*x^2 - a*c)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}{\sqrt {- c \left (a x - 1\right )} \left (a x + 1\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)**3*(-a**2*x**2+1)**(3/2)/(-a*c*x+c)**(1/2),x)

[Out]

Integral((-(a*x - 1)*(a*x + 1))**(3/2)/(sqrt(-c*(a*x - 1))*(a*x + 1)**3), x)

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Giac [A]
time = 0.42, size = 50, normalized size = 0.75 \begin {gather*} -2 \, {\left (\frac {\sqrt {a c x + c}}{a c^{2}} + \frac {2}{\sqrt {a c x + c} a c}\right )} {\left | c \right |} + \frac {4 \, \sqrt {2} {\left | c \right |}}{a c^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(1/2),x, algorithm="giac")

[Out]

-2*(sqrt(a*c*x + c)/(a*c^2) + 2/(sqrt(a*c*x + c)*a*c))*abs(c) + 4*sqrt(2)*abs(c)/(a*c^(3/2))

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Mupad [B]
time = 0.96, size = 64, normalized size = 0.96 \begin {gather*} -\frac {\left (\frac {6\,\sqrt {1-a^2\,x^2}}{a^3\,c}+\frac {2\,x\,\sqrt {1-a^2\,x^2}}{a^2\,c}\right )\,\sqrt {c-a\,c\,x}}{\frac {1}{a^2}-x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - a^2*x^2)^(3/2)/((c - a*c*x)^(1/2)*(a*x + 1)^3),x)

[Out]

-(((6*(1 - a^2*x^2)^(1/2))/(a^3*c) + (2*x*(1 - a^2*x^2)^(1/2))/(a^2*c))*(c - a*c*x)^(1/2))/(1/a^2 - x^2)

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